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ICSE Class 9 Physics Question 6 of 17

Motion in One Dimension — Question 10

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Question 10

The velocity-time graph of a moving body is given below in figure

The velocity-time graph of a moving body is shown in figure. Find The acceleration in parts AB, BC and CD, Displacement in each part AB, BC, CD, and Total displacement. Motion in one dimension, Concise Physics Solutions ICSE Class 9.

Find —

(i) The acceleration in parts AB, BC and CD.

(ii) Displacement in each part AB, BC, CD, and

(iii) Total displacement.

Answer

Let E be the point at t = 4 and F be the point at t = 8 as labelled in the graph below :

The velocity-time graph of a moving body is shown in figure. Find The acceleration in parts AB, BC and CD, Displacement in each part AB, BC, CD, and Total displacement. Motion in one dimension, Concise Physics Solutions ICSE Class 9.

(i) Acceleration in part AB = slope of AB

slope of AB=(30−0)ms−1(4−0)sslope of AB=30ms−14sslope of AB=7.5ms−2\text {slope of AB} = \dfrac{(30 - 0) \text {ms}^{-1}}{(4-0) \text {s}} \\[0.5em] \text {slope of AB} = \dfrac{30 \text {ms}^{-1}}{4 \text {s}} \\[0.5em] \text {slope of AB} = 7.5 \text {ms}^{-2} \\[0.5em]

Hence, Acceleration in part AB = 7.5 m s-2

Acceleration in part BC = slope of BC

We observe from the graph that there is no change in velocity in part BC. Hence, Acceleration in part BC = 0 m s-1

Acceleration in part CD = slope of CD

slope of CD=(0−30)ms−1(10−8)sslope of CD=−30ms−12sslope of CD=−15ms−2\text {slope of CD} = \dfrac{(0 - 30) \text {ms}^{-1}}{(10-8) \text {s}} \\[0.5em] \text {slope of CD} = \dfrac{-30 \text {ms}^{-1}}{2 \text {s}} \\[0.5em] \text {slope of CD} = -15 \text {ms}^{-2} \\[0.5em]

Hence, Acceleration in part CD = -15 m s-1

(ii) Displacement in each part is as follows —

(a) Displacement of part AB = Area of triangle ABE

=12×base×height= \dfrac {1}{2} \times \text {base} \times \text {height}

Substituting the values in the formula above, we get,

=12×(4−0)s×(30−0)m s−1=12×4s×30m s−1=30m= \dfrac {1}{2} \times {(4 -0) \text {s}} \times (30 - 0)\text {m s}^{-1} \\[0.5em] = \dfrac {1}{2} \times 4 \text {s} \times 30 \text {m s}^{-1} \\[0.5em] = 30 \text {m} \\[0.5em]

Hence,
Displacement of part AB = 60 m

(b) Displacement of part BC = Area of Square EBCF

=length×breadth= \text {length} \times \text {breadth}

Substituting the values in the formula above, we get,

=(8−4)s×(30−0)m s−1=4×30m=120m= {(8 -4) \text {s}} \times (30 - 0)\text {m s}^{-1} \\[0.5em] = 4 \times 30 \text {m} \\[0.5em] = 120 \text {m} \\[0.5em]

Hence,
Displacement of part BC = 120 m

(c) Displacement of part CD = Area of triangle CDF

=12×base×height= \dfrac {1}{2} \times \text {base} \times \text {height}

Substituting the values in the formula above, we get,

=12×(10−8)s×(30−0)m s−1=12×2×30m=30m= \dfrac {1}{2} \times {(10 -8) \text {s}} \times (30 - 0)\text {m s}^{-1} \\[0.5em] = \dfrac {1}{2} \times 2 \times 30 \text {m} \\[0.5em] = 30 \text {m} \\[0.5em]

Hence,
Displacement of part CD = 30 m

(iii) Total displacement = Displacement of part AB + Displacement of part BC + Displacement of part CD
= 60 + 30 + 120 = 210

Hence,
total displacement = 210 m